U = 0, initial vertical velocity
Neglect air resistance, and g = 9.8 m/s².
The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s
Answer: 2.0 s
To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

Where,
F = Force/Weight
A = Area
= Compression
= Original Length
According to the values given we have to




Replacing this values at our previous equation we have,



Therefore the Weight of the object is 3.82kN
Answer:
Show attached picture
Explanation:
Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call
its internal resistance) and R indicates the resistance of the light bulb.
We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:
(1)
Both the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

Using Ohm's law,
, we can rewrite the previous equation as:

where
is the current in the meter
is the current in the bulb
Using (1), this equation becomes

so, the current in the meter is 1000 times less than through the bulb.
Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
Answer:
h = v₀² / 2g
, h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²